{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 9 - Portillo Tom – Homework 9 – Due Nov 1 2006 3:00...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Portillo, Tom – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the double integral I = Z 3 2 µZ 1 x y 2 dx ¶ dy. Correct answer: 3 . 16667 . Explanation: The double integral I can be written as the repeated integral I = Z 3 2 µZ 1 x y 2 dx ¶ dy, integrating first with respect to x . Now after use of the Fundamental theorem of integral calculus, the inner integral becomes h x 2 y 2 2 i 1 = 1 2 y 2 . Thus I = Z 3 2 1 2 y 2 dy = h 1 2 ‡ y 3 3 ·i 3 2 . Consequently, I = 19 6 . keywords: 002 (part 1 of 1) 10 points Evaluate the integral I = Z 2- 1 Z 4 1 ( x 2- 2 xy ) dydx. 1. I =- 21 2 2. I =- 25 2 3. I =- 13 2 4. I =- 27 2 correct Explanation: The integral can be written in iterated form I = Z 2- 1 ‰Z 4 1 ( x 2- 2 xy ) dy ¾ dx. Now Z 4 1 ( x 2- 2 xy ) dy = h x 2 y- xy 2 i 4 1 = (4 x 2- 16 x )- ( x 2- x ) = 3 x 2- 15 x. But then I = Z 2- 1 (3 x 2- 15 x ) dx = h x 3- 15 x 2 2 i 2- 1 . Consequently, I =- 27 2 . keywords: definite integral, iterated integral, polynomial function, 003 (part 1 of 1) 10 points Evaluate the iterated integral I = Z 2 1 n Z 2 1 ‡ x y + y x · dy o dx. 1. I = 2ln 3 2 2. I = 3ln2 correct 3. I = 3ln 3 2 4. I = 3 2 ln3 5. I = 3 2 ln2 Portillo, Tom – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 2 6. I = 2ln3 Explanation: Integrating with respect to y keeping x fixed, we see that Z 2 1 µ x y + y x ¶ dy = • x ln y + y 2 2 x ‚ 2 1 = (ln2) x + 3 2 µ 1 x ¶ . Thus I = Z 2 1 • (ln2) x + 3 2 µ 1 x ¶‚ dx = •µ x 2 2 ¶ ln2 + 3 2 ln x ‚ 2 1 . Consequently, I = 3ln2 . keywords: 004 (part 1 of 1) 10 points Evaluate the double integral I = Z 3 2 Z 2 e x- y dxdy . 1. I = e- 3- e- 2- e- 1- 1 2. I = e- 3- e- 2 + e- 1 + 1 3. I = e- 3- e- 2- e- 1 + 1 correct 4. I = e- 3 + e- 2- e- 1 + 1 Explanation: After integration with respect to x , I = Z 3 2 £ e x- y / 2 dy = Z 3 2 ( e 2- y- e- y ) dy . But then, after integrating next with respect to y we see that I = £- e 2- y + e- y / 3 2 =- e- 1 + e- 3- (- 1 + e- 2 ) . Consequently, I = e- 3- e- 2- e- 1 + 1 . keywords: 005 (part 1 of 1) 10 points Determine the value of the double integral I = Z Z A 3 xy 2 4 + x 2 dA over the rectangle A = n ( x,y ) : 0 ≤ x ≤ 3 ,- 1 ≤ y ≤ 1 o , integrating first with respect to y . 1. I = ln ‡ 13 8 · 2. I = ln ‡ 4 13 · 3. I = 1 2 ln ‡ 13 8 · 4. I = 1 2 ln ‡ 13 4 · 5. I = 1 2 ln ‡ 4 13 · 6. I = ln ‡ 13 4 · correct Explanation: The double integral over the rectangle A can be represented as the iterated integral I = Z 3 µZ 1- 1 3 xy 2 4 + x 2 dy ¶ dx, Portillo, Tom – Homework 9 – Due: Nov 1 2006, 3:00 am – Inst: David Benzvi 3 integrating first with respect to y . Now after integration with respect to y with x fixed, we see that Z 1- 1 3 xy 2 4 + x 2 dy = h xy 3 4 +...
View Full Document

{[ snackBarMessage ]}

### Page1 / 11

HW 9 - Portillo Tom – Homework 9 – Due Nov 1 2006 3:00...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online